Download Algebraic degrees of stretch factors in mapping class groups 1

Algebraic degrees of stretch factors in mapping class groups
HYUNSHIK SHIN
We explicitly construct pseudo-Anosov maps on the closed surface of genus g
with orientable foliations whose stretch factor λ is a Salem number with algebraic
degree 2g. Using this result, we show that there is a pseudo-Anosov map whose
stretch factor has algebraic degree d , for each positive even integer d such that
d ≤ g.
57M50, 57M15
1
Introduction
Let Sg be a closed surface of genus g ≥ 2. The mapping class group of Sg , denoted
Mod(Sg ), is the group of isotopy classes of orientation preserving homeomorphisms
of Sg . An element f ∈ Mod(Sg ) is called a pseudo-Anosov mapping class if there
are transverse measured foliations (F u , µu ) and (F s , µs ), a number λ(f ) > 1, and a
representative homeomorphism φ such that
φ(F u , µu ) = (F u , λ(f )µu ) and φ(F s , µs ) = (F s , λ(f )−1 µs ).
In other words, φ stretches along one foliation by λ(f ) and the other by λ(f )−1 . The
number λ(f ) is called the stretch factor (or dilatation) of f .
A pseudo-Anosov mapping class is said to be orientable if its invariant foliations are
orientable. Let λH (f ) be the spectral radius of the action of f on H1 (Sg ; R). Then
λH (f ) ≤ λ(f ),
and the equality holds if and only if the invariant foliations for f are orientable (see
[5]). The number λH (f ) is called the homological stretch factor of f .
Question Which real numbers can be stretch factors?
It is a long-standing open question. Fried [4] conjectured that λ > 1 is a stretch factor
if and only if all conjugate roots of λ and 1/λ are strictly greater than 1/λ and strictly
less than λ in magnitude.
2
Hyunshik Shin
Thurston [12] showed that a stretch factor λ is an algebraic integer whose algebraic
degree has an upper bound 6g − 6. More specifically, λ is the largest root in absolute
value of a monic palindromic polynomial. Thurston gave a construction of mapping
classes of Mod(Sg ) generated by two multitwists and he mentioned that his construction
can make a pseudo-Anosov mapping class whose stretch factor has algebraic degree
6g − 6. However, he did not give specific examples.
What happens if we fix the genus g? To simplify the question, we may ask which
algebraic degrees are possible on Sg .
Question What degrees of stretch factors can occur on Sg ?
Very little is known about this question. Using Thurston’s construction, it is easy to find
quadratic integers as stretch factors. Neuwirth and Patterson [10] found non-quadratic
examples, which are algebraic integers of degree 4 and 6 on surfaces of genus 4 and
6, respectively. Using interval exchange maps, Arnoux and Yoccoz [1] gave the first
generic construction of pseudo-Anosov maps whose stretch factor has algebraic degree
g on Sg for each g ≥ 2.
Main Theorems
In this paper, we give a generic construction of pseudo-Anosov mapping classes with
stretch factor of algebraic degree 2g.
Let ci and dj be simple closed curves on Sg as in Figure 1. For k ≥ 3, let us define
fg,k = TAg,k TBg ,
k
where TAg,k = Tc1 Tc2 · · · Tcg−1 Tcg and TBg = Td1 · · · Tdg . Here, Tα is the Dehn
twist about α. We will show that fg,k is a pseudo-Anosov mapping class and its stretch
factor λ(fg,k ) is a special algebraic integer, called Salem number. A Salem number is
an algebraic integer α > 1 whose Galois conjugates other than α have absolute value
less than or equal to 1 and at least one conjugate lies on the unit circle.
Theorem A For each g ≥ 2 and k ≥ 3, fg,k is a pseudo-Anosov mapping class and
satisfies the following properties:
(1) λ(fg,k ) = λH (fg,k ),
(2) λ(fg,k ) is a Salem number, and
(3)
lim λ(fg,k ) = k − 1.
g→∞
3
Algebraic degrees of stretch factors in mapping class groups
Sg
cg
dg
c2
d2
c1
d1
...
Figure 1: Simple closed curves on Sg
In particular, we will prove that for k = 4, the algebraic degree of stretch factor is
2g. It is known that the degree of the stretch factor of a pseudo-Anosov mapping class
f ∈ Mod(Sg ) with orientable foliations is bounded above by 2g (see [12]). Therefore
our examples give the maximum degrees of stretch factors for orientable foliations in
Mod(Sg ) for each g ≥ 2.
Theorem B Let fg ∈ Mod(Sg ) be the mapping class given by
fg = fg,4 = TAg,4 TBg .
Then the minimal polynomial of the stretch factor λ(fg ) is


2g−1
X
xj  + 1.
pg (x) = x2g − 2 
j=1
This implies
deg λ(fg ) = 2g.
The hard part is to show the irreducibility of pg (x), which is proved in section 7.
In general, for each k ≥ 3, the Salem stretch factor of fg,k is the root of the polynomial


2g−1
X
pg,k (x) = x2g − (k − 2) 
xj  + 1.
j=1
It can be shown that pg,k (x) is irreducible for each k ≥ 4, but since the main purpose
of this paper is degree realization, we will prove only for k = 4 case that the algebraic
degree of the stretch factor is 2g.
Using a branched cover construction, we use Theorem B to deduce the following partial
answer to our question about algebraic degrees.
4
Hyunshik Shin
Corollary 5 For each positive integer h ≤ g/2, there is a pseudo-Anosov mapping
class feh ∈ Mod(Sg ) such that deg(λ(feh )) = 2h and λ(feh ) is a Salem number.
Obstructions.
There are three known obstructions for the existence of algebraic degrees. For any
pseudo-Anosov f ∈ Mod(Sg ), we have:
(1) deg λ(f ) ≥ 2,
(2) deg λ(f ) ≤ 6g − 6, and
(3) if deg λ(f ) > 3g − 3, then deg λ(f ) is even.
The third obstruction is due to Long [8] and we have another proof in section 5. It
turns out these are the only obstructions for g = 2. However it is not known whether
there are other obstructions of algebraic degrees for g ≥ 3. By computer search, odd
degree stretch factors are rare compared to even degrees. We conjecture that every
even degree d ≤ 6g − 6 can be realized as the algebraic degree of stretch factors.
Conjecture On Sg , there exists a pseudo-Anosov mapping class with a stretch factor
of algebraic degree d for each positive even integer d ≤ 6g − 6.
In section 6, we show that the conjecture is true for g = 2, 3, 4, and 5.
Outline
In section 2 we will give the basic definitions and results about Thurston’s consturction.
We will prove Theorem A in section 3 by the theory of Coxeter graphs. In section 4,
we construct pseudo-Anosov mapping classes via branched covers. In section 5, we
explain some properties of odd degree stretch factors. Section 6 contains examples of
even degree stretch factors for g = 2, 3, 4 and 5. Section 7 is where we prove Theorem
B, that is, we prove that the minimal polynomial of λ(fg ) has degree 2g.
Acknowledgments
I am very grateful to my advisor Dan Margalit for numerous help and discussions. I
would also like to thank Joan Birman, Benson Farb, Daniel Groves, Chris Judge, and
Balázs Strenner for helpful suggestions and comments. I wish to thank an anonymous
Algebraic degrees of stretch factors in mapping class groups
5
referee for very helpful comments. Lastly, I would like to thank the School of Mathematics of Georgia Institute of Technology for their hospitality during the time in which
the major part of this paper was made.
2
Background
Thurston’s construction
We recall Thurston’s construction of mapping classes [12]. For more details on this
material, see [3] or [6].
Suppose A = {a1 , . . . , an } is a set of pairwise disjoint simple closed curves, called a
Q
multicurve. We denote the product of Dehn twists ni=1 Tai by TA . This product is
called a multitwist.
Suppose A = {a1 , . . . , an } and B = {b1 , . . . , bm } are multicurves in a surface S so
that A ∪ B fills S, that is, the complement of A ∪ B is a disjoint union of disks and
once-punctured disks. Let N be the n × m matrix whose (j, k)-entry is the geometric
intersection number i(aj , bk ) of aj and bk . Let ν = ν(A ∪ B) be the largest eigenvalue
in magnitude of the matrix NN t . If A ∪ B is connected, then NN t is primitive and by
the Perron–Frobenius theorem ν is a positive real number greater than 1 (see [3, p.
392 - 395] for more detail).
Thurston constructed a singular Euclidean structure on S with respect to which hTA , TB i
acts by affine transformations given by the representation ρ : hTA , TB i → PSL(2, R)
1 −ν 1/2
1
0
ρ(TA ) =
and ρ(TB ) =
.
0
1
ν 1/2 1
In particular, an element f ∈ hTA , TB i is pseudo-Anosov if and only if ρ(f ) is a
hyperbolic element in PSL(2, R) and then the stretch factor λ(f ) is equal to the bigger
eigenvalue of ρ(f ). For instance, for a mapping class f = TA TB ,
1 −ν 1/2
1
0
1 − ν −ν 1/2
ρ(TA TB ) =
=
,
0
1
ν 1/2 1
ν 1/2
1
and the stretch factor λ(TA TB ) is the bigger root of the characteristic polynomial
λ2 − λ(ν − 2) + 1,
provided that ν − 2 > 2.
6
Hyunshik Shin
...
Figure 2: Multicurves and configuration graph G(Ag,k ∪ Bg )
3
Proof by the theory of Coxeter graphs
We will prove Theorem A in this section.
For the set C of simple closed curves on the surface Sg , the configuration graph for
C, denoted G(C), is the graph with a vertex for each simple closed curve and an edge
for every point of intersection between simple closed curves.
Let fg,k be a mapping class on Sg defined by
fg,k = TAg,k TBg , k ≥ 3,
as in Theorem A. By regarding the multiple power of Tcg as the product of Dehn twists
about parallel (isotopic) simple closed curves cg1 , . . . , cgk , let us define the multicurves
Ag,k = {c1 , . . . , cg−1 , cg1 , . . . , cgk } and Bg = {d1 , . . . , dg }.
Then the configuration graph G(Ag,k ∪ Bg ) is a tree as in Figure 2,
3.1
Coxeter graphs and mapping class groups
We say that a finite graph G is a Coxeter graph if there are no self-loops or multiple
edges. For given multicurves A and B such that A ∪ B fills the surface S, suppose
that the configuration graph G = G(A ∪ B) is a Coxeter graph. Leininger proved the
following theorem.
Algebraic degrees of stretch factors in mapping class groups
7
Theorem 1 ([6] Theorem 8.1 and Theorem 8.4) Let G(A ∪ B) be a non-critical
dominant Coxeter graph. Then TA TB is a pseudo-Anosov mapping class with stretch
factor λ such that
λ2 + λ(2 − µ2 ) + 1 = 0,
where µ is the spectral radius of the graph G .
For the definitions and pictures of critical and dominant graphs, see [6, Section 1]
For the multicurves Ag,k and Bg in Theorem A, G(Ag,k ∪ Bg ) is a non-critical dominant
Coxeter graph for each k ≥ 3. Therefore by Theorem 1 the mapping class fg,k = TA TB
is pseudo-Anosov for each k ≥ 3.
3.2
Orientability
Suppose that G is a connected Coxeter graph with the set Σ of vertices. There is an
associated quadratic form ΠG on RΣ and a faithful representation
Θ : C (G) → O(ΠG ),
where C (G) is a Coxeter group with generating set Σ, O(ΠG ) is the orthogonal group
of the quadratic form ΠG , and each generator si ∈ Σ is represented by a reflection.
Leininger also proved the following theorem.
Theorem 2 ([6] Theorem 8.2 ) Let G(A ∪ B) be a Coxeter graph and suppose that
A and B can be oriented so that all intersections of A with B are positive. Then there
exists a homomorphism
η : RΣ → H1 (S; R)
such that
(TA TB )∗ ◦ η = −η ◦ Θ(σA σB ),
where σA σB is an element in C (G) corresponding to TA TB .
Moreover, Θ(σA σB )|ker(η) = −I and η preserves spectral radii.
Theorem 2 implies that if A and B can be oriented as in the theorem, then the stretch
factor of a pseudo-Anosov mapping class is equal to the spectral radius of the action
on homology. For multicurves Ag,k and Bg in Theorem A, they can be oriented so that
all intersections are positive as in Figure 3. Therefore we have
λ(fg,k ) = λH (fg,k )
8
Hyunshik Shin
...
Figure 3: Orientation of positive intersections
and the invariant foliations for fg,k are orientable.
It is also possible to directly compute the action on the first homology. Consider the
mapping class fg = TAg,4 TBg as in Theorem B. Let us choose a basis {a1 , b1 , . . . , ag , bg }
for H1 (Sg ) as in Figure 4.
Sg
bg
ag
a3
b2
a2
b1
a1
...
Figure 4: A basis for H1 (Sg ).
By computing images of each basis element under fg , we can get the action on H1 (Sg )


1 −1
0 0 ···
0
 1
0 −1 0 · · ·
0 


 ..
..
.. .. . .
..  .
 .

.
.
.
.
.


 1
0
0 0 · · · −1 
4
0
0
0 ···
−3
By induction, the characteristic polynomial hg (x) of the homological action is


2g−1
X
hg (x) = x2g + 2 
(−1)j xj  + 1.
j=1
Algebraic degrees of stretch factors in mapping class groups
9
Since the largest root of hg (x) in magnitude is a negative real number, we can deduce
that the stretch factor λ(fg ) is the root of hg (−x). Specifically, λ(fg ) is the root of


2g−1
X
pg (x) = x2g − 2 
xj  + 1.
j=1
In a similar way, one can get the polynomial for λ(fg,k ), which is


2g−1
X
pg,k (x) = x2g − (k − 2) 
xj  + 1.
j=1
3.3
Salem numbers and spectral properties of starlike trees
The configuration graph G(Ag,k ∪Bg ) for fg,k is a special type of graphs, called a starlike
tree, and its relation to Salem numbers is studied in [9]. A starlike tree is a tree with at
most one vertex of degree > 2. Let T = T(n1 , n2 , . . . , nk ) be the starlike tree with k
arms of n1 , n2 , . . . , nk edges.
Theorem 3 ([9] Corollary 9) Let T = T(n1 , n2 , . . . , nk ) be a starlike tree and let µ
be the spectral radius of T . Suppose that√µ is not√an integer and T is a non-critical
dominant graph. Then λ > 1, defined by λ + 1/ λ = µ, is a Salem number.
The configuration graph G(Ag,k ∪ Bg ) in Theorem A is a non-critical dominant starlike
tree
T(2g − 2, 1, 1, . . . , 1), k ≥ 3
| {z }
k-times
and we will denote it by T(2g−2, k·1). The fact that the spectral radius of T(2g−2, k·1)
is not an integer follows from the following theorem.
Theorem 4 ([11]) If µ is the spectral radius of the starlike tree T(n, k · 1), then
√
for n ≥ 1 and k ≥ 3.
k
k+1<µ< √
k−1
10
Hyunshik Shin
Thus for the starlike tree T(n, k · 1), the spectral radius satisfies
k + 1 < µ2 <
k2
1
=k+1+
.
k−1
k−1
Therefore µ is not an integer and by Theorem 3 λ(fg,k ) is a Salem number.
Moreover, the proof of Corollary 2.1 of Lepović–Gutman [7] implies that
lim λ(fg,k ) = k − 1.
g→∞
For completeness, we reprove this here.
Recall that λ(fg,k ) is the largest root of

2g−1
X
pg,k (x) = x2g − (k − 2) 

xj  + 1.
j=1
By multiplying pg,k (x) by x−1, the stretch factor λ(fg,k ) is the largest root in magnitude
of
qg,k (x) = x2g+1 − (k − 1)x2g + (k − 1)x − 1.
We have qg,k (k − 1) = (k − 1)2 − 1 > 0, and for any fixed positive integer m,
1
1 2g
1
1
qg,k k − 1 − m = k − 1 − m
− m + (k − 1) k − 1 − m − 1
10
10
10
10
Hence qg,k k − 1 − 10−m < 0 for sufficiently large values of g and therefore pg,k (x)
has a root on the interval (k − 1 − 10−m , k − 1). This implies
lim λ(fg,k ) = k − 1.
g→∞
This completes the proof of Theorem A.
Remark A positive integer cannot be a stretch factor (which is an algebraic integer
of degree 1). However, Theorem A implies that for sufficiently large genus g there is
a stretch factor which is a Salem number arbitrarily close to a given integer k − 1 for
each k ≥ 3.
11
Algebraic degrees of stretch factors in mapping class groups
4
Branched Covers
Lifting a pseudo-Anosov mapping class via a covering map is one way to construct
another pseudo-Anosov mapping class. If there is a branched cover e
S → S and a
pseudo-Anosov mapping class f ∈ Mod(S), then there is some k ∈ N such that
Mod(e
S) has a pseudo-Anosov element ef which is a lift of f k and hence λ(ef ) = λ(f )k .
Corollary 5 Let g ≥ 2. For each positive integer h ≤ g/2, there is a pseudo-Anosov
mapping class feh ∈ Mod(Sg ) such that deg(λ(feh )) = 2h and λ(feh ) is a Salem number.
Proof Let
h=


g−2m
2 ,
if g is even, m = 0, 1, . . . , (g − 2)/2,

g−1−2m
,
2
if g is odd, m = 0, 1, . . . , (g − 3)/2.
Then h is an integer such that 1 ≤ h ≤ g/2.
Sg
Sg
Sh
Sh
g odd
g even
Figure 5: A branched cover
Construct a branched cover Sg → Sh as in Figure 5. For h ≥ 2, Sh has a pseudoAnosov mapping class fh ∈ Mod(Sh ) as in the Theorem B whose stretch factor has
12
Hyunshik Shin
deg(λ(fh )) = 2h. For some
k, fh k lifts to Sg and the lift has stretch factor λ(fh )k . We
claim that deg λ(fh )k = 2h. To see this, let λi , 1 ≤ i ≤ 2h, be the roots of the
minimal polynomial of λ(fh ) and let us define a polynomial
p(x) =
2h
Y
x − λki .
i=1
Then p(x) is an integral polynomial because the following elementary symmetric
polynomials in λ1 , . . . , λ2h
X
X
X
λi ,
λi λj ,
λi λj λl , · · · , λ1 λ2 · · · λ2h
i<j
i<j<l
are all integers and hence each coefficient of p(x)
X
X
X
λki ,
λki λkj ,
λki λkj λkl , · · · , λk1 λk2 · · · λk2h
i<j
i<j<l
are integers as well. Therefore p(x) is divided by the minimal polynomial of λ(fh )k .
Due to the proof of Theorem B in section 7, λ(fh )k is also a Salem number and
p(x) does not
have a cyclotomic factor. This implies that p(x) is irreducible and
k
deg λ(fh ) = 2h.
If h = 1, Sh is a torus and it admits a Anosov mapping class f whose stretch factor
λ(f ) has algebraic degree 2. Then similar arguments as above tells us that there is a
lift of some power of f to Sg whose stretch factor has deg(λ(f k )) = 2.
Therefore there is a pseudo-Anosov map feh ∈ Mod(Sg ) with deg(λ(feh )) = 2h for each
h ≤ g/2. In other words, every positive even degree d ≤ g is realized as the algebraic
degree of a stretch factor on Sg .
5
Stretch factors of odd degrees
Long proved the following degree obstruction and McMullen communicated to us the
following proof. First we will give a definition of the reciprocal polynomial. Given
a polynomial p(x) of degree d , we define the reciprocal polynomial p∗ (x) of p(x) by
p∗ (x) = xd p(1/x). It is a well-known property that p∗ (x) is irreducible if and only if
p(x) is irreducible.
Theorem 6 ([8]) Let f ∈ Mod(Sg ) be a pseudo-Anosov mapping class having stretch
factor λ(f ). If deg(λ(f )) > 3g − 3, then deg(λ(f )) is even.
Algebraic degrees of stretch factors in mapping class groups
13
Proof Since f acts by a piecewise integral projective transformation on the 6g − 6
dimensional space PMF of projective measured foliations on Sg , and since λ(f ) is an
eigenvalue of this action, λ(f ) is an algebraic integer with deg(λ(f )) ≤ 6g − 6. Also,
since f preserves the symplectic structure on PMF , it follows that λ(f ) is the root of
palindromic polynomial p(x) whose degree is bounded above by 6g − 6.
Let q(x) be the minimal polynomial of λ(f ) and let q∗ (x) be the reciprocal polynomial
of q(x). Then either q(x) = q∗ (x) or they have no common roots, because if there
is at least one common root ζ of q(x) and q∗ (x), then both q(x) and q∗ (x) are the
minimal polynomial of ζ and hence q(x) = q∗ (x). Suppose deg(q(x)) > 3g − 3.
If q(x) and q∗ (x) have no common roots, then their product q(x) q∗ (x) is a factor of
p(x) since q∗ (x) is the minimal polynomial
of 1/λ(f ). This is a contradiction because
deg(p(x)) ≤ 6g−6 but deg q(x) q∗ (x) > 6g−6. Therefore we must have q(x) = q∗ (x)
and this implies that q(x) is an irreducible palindromic polynomial. Hence deg(q(x))
is even since roots of q(x) come in pairs, λi and 1/λi .
It follows from the previous proof that if the minimal polynomial p(x) of λ has odd
degree, then p(x) is not palindromic and in fact the minimal palindromic polynomial
containing λ as a root is p(x)p∗ (x).
We will now show that the stretch factors of degree 3 have an additional special property.
A Pisot number, also called a Pisot–Vijayaraghavan number or a PV number, is an
algebraic integer greater than 1 such that all its Galois conjugates are strictly less than
1 in absolute value.
Proposition 7 Let f ∈ Mod(Sg ). If deg(λ(f )) = 3, then λ(f ) is a Pisot number.
Proof Let λ1 > 1 be the stretch factor of a pseudo-Anosov mapping class with
algebraic degree 3, and let p(x) be the minimal polynomial of λ1 . Let λ1 , λ2 , and
λ3 be the roots of p(x). Then the degree of p(x)p∗ (x) is 6 and it has pairs of roots
(λ1 , 1/λ1 ), (λ2 , 1/λ2 ), (λ3 , 1/λ3 ), where λ1 is the largest root in absolute value. We
claim that the absolute values of λ2 and λ3 are strictly less than 1.
Suppose one of them has absolute value greater than or equal to 1, say |λ2 | ≥ 1. The
constant term λ1 λ2 λ3 of p(x) is ±1 since it is the factor of a palindromic polynomial
with constant term 1. Hence |λ1 λ2 λ3 | = 1 and we have
1
= |λ1 λ2 | ≥ |λ1 |,
|λ3 |
which is a contradiction to the fact that the stretch factor λ1 is strictly greater than all
other roots of the palindromic polynomial p(x)p∗ (x). This proves the claim and hence
the stretch factor of degree 3 is a Pisot number.
14
Hyunshik Shin
We now explain two constructions of mapping classes f ∈ Mod(Sg ) whose degree of
λ(f ) is odd.
1. As we mentioned, Arnoux–Yoccoz [1] gave examples of a pseudo-Anosov mapping
class on Sg whose stretch factor has algebraic degree g. In particular for odd g, this
gives examples of mapping classes with odd degree stretch factors. They proved that
these stretch factors are all Pisot numbers.
2. For genus 2, there is a pseudo-Anosov mapping class f whose stretch factor has
algebraic degree 3 (see section 6). This is the only possible odd degree on S2 by Long’s
obstruction. It is also true that deg(λ(f )k ) = 3 for each k because the stretch factor is a
Pisot number (Proposition 7). There is a cover Sg → S2 for each g, so the lift of some
power of f has a stretch factor with algebraic degree 3 on Sg .
Proposition 8 For each genus g, the stretch factor with algebraic degree 3 can occur
on Sg .
Question Are there stretch factors with odd algebraic degree that are not Pisot numbers?
6
Examples of even degrees
Tables 1 through 4 give explicit examples of pseudo-Anosov mapping classes whose
stretch factors realize various degrees. We will follow the notation of the software
Xtrain by Brinkmann. More specifically, ai , bi , ci , and di are Dehn twists along
standard curves and Ai , Bi , Ci , and Di are the inverse twists as in [2]. The only missing
degree on S3 is degree 5. We do not know if there is a degree 5 example or there is
another degree obstruction.
deg
f ∈ Mod(S2 )
2
a0 a0 d0 C0 D1 C0
x2 − 3x + 1
λ = 2.618
a0 d0 d0 C0 C0 D1
x3
λ = 3.383
4
a0 d0 d0 d1 c0 d0
x4
6
a0 a0 d0 A0 C0 D1
x6 − x5 − 4x3 − x + 1
3
Minimal polynomial
−
3x2
−
x3
−x−1
−
x2
−x+1
Table 1: Examples of genus 2
λ(f )
λ = 1.722
λ = 2.015
15
Algebraic degrees of stretch factors in mapping class groups
f ∈ Mod(S3 )
deg
2
3
4
a1 c0 d0 c0 d2 C1 D1
a0 c0 d0 C1 D1 D2
a1 c0 d0 a1 c1 d1 d2
3.732
x − 4x + 1
3
2
1.755
x − 2x + x − 1
4
3
2
1.722
x − x − 2x − x + 1
6
5
4
3
2
6
a0 c0 d0 d2 C1 D1
x − 3x + 3x − 7x + 3x − 3x + 1
8
a0 c0 d0 d1 C1 D2
x8 − x7 − 2x5 − 2x3 − x + 1
10
12
a1 c0 d0 d1 C1 A2 D2
a1 c1 c0 d1 d2 A0 D0
λ(f )
Minimal polynomial
2
10
9
8
12
11
9
7
2.739
1.809
5
3
2
1.697
3
1.533
x − x − 2x + 2x − 2x + 2x − 2x − x + 1
8
7
5
4
x −x −x −x +x +x −x −x −x+1
Table 2: Examples of genus 3
deg
f ∈ Mod(S4 )
deg
f ∈ Mod(S4 )
4
a0 a0 a1 c0 d0 c1 d1 c2 d2 c3 d3
12
a0 B1 d0 c0 d1 c1 d2 c2 d3 c3
6
a0 B2 A3 d0 c0 d1 c1 d2 c2 d3 c3
14
a0 d0 B0 d0 c0 d1 c1 d2 c2 d3 c3
8
a0 A1 d0 c0 d1 c1 d2 c2 d3 c3
16
A0 d0 c0 d1 c1 d2 c2 d3 c3
10
a0 b1 A2 d0 c0 d1 c1 d2 c2 d3 c3
18
a0 B1 A2 d0 c0 d1 c1 d2 c2 d3 c3
Table 3: Examples of genus 4
deg
f ∈ Mod(S5 )
deg
f ∈ Mod(S5 )
6
b3 d0 c0 d1 c1 d2 c2 d3 c3 d4 c4
16
a1 B2 d0 c0 d1 c1 d2 c2 d3 c3 d4 c4
8
a0 a1 d0 c0 d1 c1 d2 c2 d3 c3 d4 c4
18
a1 B0 d0 c0 d1 c1 d2 c2 d3 c3 d4 c4
10
a1 A4 d0 c0 d1 c1 d2 c2 d3 c3 d4 c4
20
a1 A0 d0 c0 d1 c1 d2 c2 d3 c3 d4 c4
12
b2 C2 d0 c0 d1 c1 d2 c2 d3 c3 d4 c4
22
a2 A1 d0 c0 d1 c1 d2 c2 d3 c3 d4 c4
14
a1 B1 d0 c0 d1 c1 d2 c2 d3 c3 d4 c4
24
c2 A2 d0 c0 d1 c1 d2 c2 d3 c3 d4 c4
Table 4: Examples of genus 5
7
Irreducibility of Polynomials
In this section, we will prove Theorem B. It is enough to show that the polynomial


2n−1
X
pn (x) = x2n − 2 
xj  + 1.
j=1
is irreiducible for n ≥ 2. We will show that pn (x) does not have a cyclotomic
polynomial factor. It then follows from Kronecker’s theorem that pn (x) is irreducible.
16
Hyunshik Shin
Suppose pn (x) has the mth cyclotomic polynomial factor for some m ∈ N. Then
e2πi/m is a root of pn (x). Multiplying pn (x) by x − 1 yields
x2n+1 − 3x2n + 3x − 1
and hence we have
(1)
e2(2n+1)πi/m − 3e4nπi/m + 3e2πi/m − 1 = 0.
Consider the real part and the complex part of (1). Then we have the system of
equations
− 3 cos 4nπ
+ 3 cos 2π
−1=0
cos 2(2n+1)π
m
m
m
2(2n+1)π
4nπ
2π
sin
− 3 sin m + 3 sin m = 0
m
Using double-angle formula for the first cosine and sum-to-product formula for the last
two cosines, the first equation gives
(2n + 1)π h
(2n − 1)π (2n + 1)π i
2 sin
3 sin
− sin
= 0.
m
m
m
Similarly the second equation gives
(2n + 1)π h
(2n + 1)π (2n − 1)π i
2 cos
sin
− 3 sin
= 0.
m
m
m
Since sine and cosine have no common zeros, we must have
(2n − 1)π (2n + 1)π − 3 sin
= 0.
sin
m
m
For m ≤ 5, by direct calculation we can see that pn (e2πi/m ) 6= 0. So we may assume
that m ≥ 6. Let ϕ = (2n − 1)π/m and then we can write the above equation as
2π (2)
sin ϕ +
− 3 sin(ϕ) = 0.
m
Since sin ϕ + 2π/m is a real number between −1 and 1, we have
(3)
−
1
1
≤ sin(ϕ) ≤ .
3
3
Let ψ = sin−1 (1/3). Then note that ψ < π/6. Equation (3) gives the restriction on
ϕ, which is
−ψ ≤ ϕ ≤ ψ or π − ψ ≤ ϕ ≤ π + ψ.
Another observation from (2) is that both sin ϕ + 2π/m and sin(ϕ) must have the
same sign.
We claim that ϕ has to be on the either first or third quadrant. Suppose ϕ is on the
second quadrant, that is, π − ψ < ϕ < π . Note that m ≥ 6 implies 2π/m ≤ π/3.
Algebraic degrees of stretch factors in mapping class groups
17
Since ϕ is above the x-axis, ϕ + 2π/m also has to be above the x-axis due to (2) and
hence the only possibility is that ϕ + 2π/m is between ϕ and π . Then
2π 2π 0 < sin ϕ +
< sin(ϕ) =⇒ sin ϕ +
< 3 sin(ϕ),
m
m
which is a contradiction to (2). Similar arguments hold if ϕ is on the fourth quadrant.
Therefore the possible range for ϕ is
0 < ϕ ≤ ψ or π < ϕ ≤ π + ψ.
Suppose ϕ is on the first quadrant. Then so is ϕ + 2π/m because
0<ϕ+
We can write
ϕ=
2π
π
π
≤ψ+ < .
m
3
2
(2n − 1)π
jπ
≡
m
m
(mod 2π)
for some positive integer j, i.e., 0 < jπ/m < π/2.
If j ≥ 2, Using the subadditivity of sin(x) on the first quadrant
sin(x + y) ≤ sin(x) + sin(y),
we have
2π sin(ϕ) + sin
− 3 sin(ϕ)
m
2π = sin
− 2 sin(ϕ)
m
2π jπ = sin
− 2 sin
< 0,
m
m
2π sin ϕ +
− 3 sin(ϕ) ≤
m
which contradicts (2).
If j = 1, using triple-angle formula
3π π
2π sin ϕ +
− 3 sin(ϕ) = sin
− 3 sin
m
m m π
π
π
= 3 sin
− 4 sin3
− 3 sin
m
m
m 3 π
= −4 sin
< 0,
m
which contradicts (2) again. Therefore there is no possible ϕ on the first quadrant. By
using the same arguments, the fact that ϕ is on the third quadrant gives a contradiction.
Therefore we can conclude that p(x) does not have a cyclotomic factor.
18
Hyunshik Shin
We now show that pn (x) is irreducible over Z. Suppose pn (x) is reducible and write
pn (x) = g(x)h(x) with non-constant functions g(x) and h(x). There is only one root
of pn (x) whose absolute value is strictly greater than 1. Therefore one of g(x) or h(x)
has all roots inside the unit disk. By Kronecker’s theorem, this polynomial has to be
a product of cyclotomic polynomials, which is a contradiction because pn (x) does not
have a cyclotomic polynomial factor. Therefore pn (x) is irreducible.
References
[1] Pierre Arnoux and Jean-Christophe Yoccoz. Construction de difféomorphismes pseudoAnosov. C. R. Acad. Sci. Paris Sér. I Math., 292(1):75–78, 1981.
[2] Peter Brinkmann. An implementation of the Bestvina-Handel algorithm for surface
homeomorphisms. Experiment. Math., 9(2):235–240, 2000.
[3] Benson Farb and Dan Margalit. A primer on mapping class groups, volume 49 of
Princeton Mathematical Series. Princeton University Press, Princeton, NJ, 2012.
[4] David Fried. Growth rate of surface homeomorphisms and flow equivalence. Ergodic
Theory Dynam. Systems, 5(4):539–563, 1985.
[5] Erwan Lanneau and Jean-Luc Thiffeault. On the minimum dilatation of pseudoAnosov homeromorphisms on surfaces of small genus. Ann. Inst. Fourier (Grenoble),
61(1):105–144, 2011.
[6] Christopher J. Leininger. On groups generated by two positive multi-twists: Teichmüller curves and Lehmer’s number. Geom. Topol., 8:1301–1359 (electronic),
2004.
[7] M. Lepović and I. Gutman. Some spectral properties of starlike trees. Bull. Cl. Sci.
Math. Nat. Sci. Math., (26):107–113, 2001. The 100th anniversary of the birthday of
Academician Jovan Karamata.
[8] D. D. Long. Constructing pseudo-Anosov maps. In Knot theory and manifolds (Vancouver, B.C., 1983), volume 1144 of Lecture Notes in Math., pages 108–114. Springer,
Berlin, 1985.
[9] J. F. McKee, P. Rowlinson, and C. J. Smyth. Salem numbers and Pisot numbers
from stars. In Number theory in progress, Vol. 1 (Zakopane-Kościelisko, 1997), pages
309–319. de Gruyter, Berlin, 1999.
[10] L. Neuwirth and N. Patterson. A sequence of pseudo-Anosov diffeomorphisms. In
Combinatorial group theory and topology (Alta, Utah, 1984), volume 111 of Ann. of
Math. Stud., pages 443–449. Princeton Univ. Press, Princeton, NJ, 1987.
[11] Hyunshik Shin. Spectral radius of starlike trees with one long arm. Preprint.
Algebraic degrees of stretch factors in mapping class groups
19
[12] William P. Thurston. On the geometry and dynamics of diffeomorphisms of surfaces.
Bull. Amer. Math. Soc. (N.S.), 19(2):417–431, 1988.
Department of Mathematical Sciences, KAIST
291 Daehak-ro Yuseong-gu Daejeon 34141 South Korea
[email protected]
http://mathsci.kaist.ac.kr/~hshin